The median of a random variable X is a number µ that satisfies

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Find the median of the exponential random variable with parameter λ.

  • $\begingroup$ Hint: write first hte CDF of exponential law... $\endgroup$ – Jean-Claude Arbaut Mar 23 '13 at 16:14
  • $\begingroup$ Sorry but... what kind of approach did you try, which failed to lead you to the solution? $\endgroup$ – Did Mar 23 '13 at 17:03

For any positive number $x$, we have $$F_X(x)=\Pr(X\le x)=\int_0^x \lambda e^{-\lambda t}\,dt=1-e^{-\lambda x}.$$

So to find the median $m$, we want to solve the equation $$1-e^{-\lambda m}=\frac{1}{2}$$ for $m$, or equivalently to solve the equation $$e^{-\lambda m}=\frac{1}{2}.$$

Take the (natural) logarithm of both sides. We get $$-\lambda m=\ln(1/2)=-\ln(2).$$ Now solve for $m$. We get $$m=\frac{\ln 2}{\lambda}.$$

Remark: It is unfortunate that in the problem the symbol $\mu$ is used for the median, since $\mu$ is a standard name for the mean. A common symbol for the median is $m$.

  • $\begingroup$ Maybe the author's intention was to declare the rate as 1/2 and meant to use "λ" rather than "μ". $\endgroup$ – Tommyixi Sep 16 '16 at 0:50
  • $\begingroup$ @Tommyixi: Sorry, still recovering, will not be able to respond for a few days. $\endgroup$ – André Nicolas Sep 23 '16 at 23:35

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